By Steven Dale Cutkosky
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Additional resources for Class notes on linear algebra [Lecture notes]
Example text
Then the map ϕz : V → F defined by ϕz (v) =< v, z > is a linear map. Suppose that ψ ∈ V ∗ . Then there exists a unique z ∈ V such that ψ = ϕz . Proof. Let {v1 , . . , vn } be an orthonormal basis of V with dual basis {v1∗ , . . , vn∗ }. Suppose that ψ ∈ V ∗ . Then there exist (unique) c1 , . . , cn ∈ F such that ψ = c1 v1∗ + · · · + cn vn∗ . Let z = c1 v1 + · · · + cn vn ∈ V . We have that ψ(vi ) = ci for 1 ≤ i ≤ n. ϕz (vi ) =< vi , z >= c1 < vi , v1 > + · · · + cn < vi , vn >= ci = ψ(vi ) 31 for 1 ≤ i ≤ n.
V, L∗ (cw) >=< L(v), cw >= c < L(v), w >= c < v, L∗ (w) >=< v, cL∗ (w) > . Since this identity holds for all v ∈ V , we have that L∗ (cw) = cL∗ (w). Thus L∗ is linear. Now we prove uniqueness. Suppose that T : V → V is a linear map such that < L(v), w >=< v, T (w) > for all v, w ∈ V . Then for any w ∈ W , < v, T (w) >=< v, L∗ (w) > for all v ∈ V . Thus T (w) = L∗ (w), and L∗ is unique. L∗ is called the adjoint of L. 3. L is called self adjoint if L∗ = L. If V is a real inner product space, a self adjoint operator is called symmetric, and we sometimes write L∗ = Lt .
For all w ∈ W we have < Lv, w >=< v, L∗ w >=< v, Lw >= 0, since Lw ∈ W . Hence Lv ∈ W ⊥ . 10. Suppose that V is an Hermitian inner product space and L is Hermitian. Suppose that v ∈ V . Then < Lv, v >∈ R. Proof. We have < Lv, v >=< v, L∗ v >=< v, Lv >= < Lv, v >. Thus < Lv, v >∈ R. 21. 1. Let A ∈ Mnn (R) be symmetric. Then A has a real eigenvalue. Proof. Regarding A as a matrix in Mn,n (C), we have that A is Hermitian. Now A has a complex eigenvalue λ, with an eigenvector v ∈ Cn . Let < , > be the standard Hermitian t product on Cn .